AND xor OR


Problem Statement :


Given an array  of  distinct elements. Let  and  be the smallest and the next smallest element in the interval  where .

.

where , are the bitwise operators ,  and  respectively.
Your task is to find the maximum possible value of .

Input Format

First line contains integer N.
Second line contains N integers, representing elements of the array A[] .


Output Format

Print the value of maximum possible value of Si.



Solution :


                            Solution in C :

In C ++ :





#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;


int main() {
   
    
    
    int n;
    cin>>n;
    
    int * arr;
    
    
    arr =  new int[n];
    
    
    for(int i=0;i<n;i++)cin>>arr[i];
    
    stack<int> S;
    
    int m = 0;
    
    for(int i=0;i<n;i++)
    {
        while(true)
        {
            if(S.empty())
            {
                S.push(arr[i]);
                break;
            }
            
            else if(S.top() < arr[i] )
            {
                m = max(m, S.top()^arr[i]);
                S.push(arr[i]);
                break;
            }
            else
            {
                S.pop();
            }
        }        
    }
    
    
    stack<int> S2;
    
    
    for(int i=n-1;i>=0;i--)
    {
        while(true)
        {
            if(S.empty())
            {
                S.push(arr[i]);
                break;
            }
            
            else if(S.top() < arr[i] )
            {
                m = max(m, S.top()^arr[i]);
                S.push(arr[i]);
                break;
            }
            else
            {
                S.pop();
            }
        }        
    }
    
    
    
    cout<<m<<endl;
    return 0;
}








In Java :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++){
            arr[i] = in.nextInt();
        }
        
        Stack<Integer> stack = new Stack<Integer>();
        int result = 0;
        for(int i = 0; i < n; i++){
            while(true){
                if(stack.isEmpty()){
                    stack.add(arr[i]);
                    break;
                } else if(stack.peek() < arr[i]){
                    result = Math.max(result, stack.peek() ^ arr[i]);
                    stack.add(arr[i]);
                    break;
                } else{
                    stack.pop();
                }
            }
        }
        stack = new Stack<Integer>();
        
        for(int i = n - 1; i >= 0; i--){
            while(true){
                if(stack.isEmpty()){
                    stack.add(arr[i]);
                    break;
                } else if(stack.peek() < arr[i]){
                    result = Math.max(result, stack.peek() ^ arr[i]);
                    stack.add(arr[i]);
                    break;
                } else{
                    stack.pop();
                }
            }
        }
        
        System.out.println(result);        
    }
}








In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

long long int sol(int A, int B)
{
	return (((A & B) ^ (A | B)) & (A ^ B));
}

long long int inter(long long int arr[], int n)
{
	int i,j;
	long long int min1 = 9999999999, min2 = 9999999999, max = -1;
	for(i=0;i<n-1;i++)
	{
			if(sol(arr[i], arr[i+1]) > max)
				max = sol(arr[i], arr[i+1]);
        if(i+2 <= n-1)
            if(arr[i+1] > arr[i] && arr[i+1] > arr[i+2] && sol(arr[i], arr[i+2]) > max)
                max = sol(arr[i], arr[i+2]);
		
	}
	return max;
	
}

int main()
{
    int n,i;
	scanf("%d", &n);
	long long int arr[n];
	for(i=0;i<n;i++)
	{
		scanf("%lli", &arr[i]);
	}
		printf("%lli", inter(arr, n));
    return 0;
}








In  Python3 :





n_elements = int(input())
data = list(map(int, input().split()))
assert len(data) == n_elements

res = 0
minima = [data[0]]
for i in range(1, n_elements):
    d = data[i]
    while len(minima):
        x = minima.pop()
        res = max(d ^ x, res)
        if x < d:
            minima.append(x)
            break
    minima.append(d)
print(res)
                        




View More Similar Problems

Mr. X and His Shots

A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M

View Solution →

Jim and the Skyscrapers

Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space

View Solution →

Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

View Solution →

Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

View Solution →

Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →

Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →